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\(n_{CuO}=\dfrac{3,2}{80}=0,04\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{100.20\%}{98}=0,204\left(mol\right)\)
PTHH:
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
0,04 0,04 0,04
\(\dfrac{0,04}{1}< \dfrac{0,204}{1}\) --> H2SO4 dư
\(C\%_{CuSO_4}=\dfrac{0,04.160}{3,2+100}.100\%=6,2\%\)
\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,2.98}{3,2+100}.100\%=19\%\)
\(a)Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ b)Đặt:n_{Zn}=x\left(mol\right);n_{Fe}=y\left(mol\right)\\ Tacó:\left\{{}\begin{matrix}65x+56y=12,1\\x+y=0,2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\\ \Rightarrow m_{Fe}=0,1.56=5,6\left(g\right);m_{Zn}=0,1.65=6,5\left(g\right)\\ c)n_{H_2SO_4}=n_{H_2}=0,2\left(mol\right)\\ \Rightarrow C\%_{H_2SO_{\text{ 4}}}=\dfrac{0,2.98}{196}.100=10\%\)
\(n_{CuO}=\dfrac{1,6}{80}=0,02\left(mol\right)\)
\(m_{H_2SO_4}=100.20\%=20\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{20}{98}=\dfrac{10}{49}\left(mol\right)\)
PTHH: CuO + H2SO4 → CuSO4 + H2O
Mol: 0,02 0,02 0,02
Ta có: \(\dfrac{0,02}{1}< \dfrac{\dfrac{10}{49}}{1}\) ⇒ CuO hết, H2SO4 dư
mdd sau pứ = 1,6 + 100 = 101,6 (g)
\(C\%_{ddCuSO_4}=\dfrac{0,02.160.100\%}{101,6}=3,15\%\)
\(C\%_{ddH_2SO_4}=\dfrac{\left(\dfrac{10}{49}-0,02\right).98.100\%}{101,6}=17,76\%\)
$n_{Na_2CO_3} = \dfrac{265.10\%}{106} = 0,25(mol)$
$n_{CaCl_2} = \dfrac{475,72.7\%}{111} = 0,3(mol)$
$CaCl_2 + Na_2CO_3 \to CaCO_3 + 2NaCl$
Ta thấy : $n_{CaCl_2} > n_{Na_2CO_3}$ nên $CaCl_2$ dư
$n_{CaCl_2\ dư} = 0,3 - 0,25 = 0,05(mol)$
$n_{NaCl} = 0,5(mol)$
Sau phản ứng, $m_{dd} = 265 + 475,72 - 0,25.100 = 715,72(gam)$
$C\%_{NaCl} = \dfrac{0,5.58,5}{715,72}.100\% = 4,09\%$
$C\%_{CaCl_2\ dư} = \dfrac{0,05.111}{715,72}.100\% = 0,76\%$
a)
$n_{Al} = \dfrac{5,4}{27} = 0,2(mol)$
$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
Theo PTHH:
$n_{H_2SO_4} = n_{H_2} = \dfrac{3}{2}n_{Al} = 0,3(mol)$
$V_{dd\ H_2SO_4} = \dfrac{0,3}{2} = 0,15(lít)$
$n_{Al_2(SO_4)_3} = \dfrac{1}{2}n_{Al} = 0,1(mol)$
$C_{M_{Al_2(SO_4)_3}} = \dfrac{0,1}{0,15} = 0,67M$
b)
$V_{H_2} = 0,3.22,4 = 6,72(lít)$
a/ PTHH : CuO + H2SO4 ===> CuSO4 + H2O
0,04 0,04 0,04 ( mol )
b/ mH2SO4= 150 x 10% = 15 gam
=> nH2SO4= 15 : 98 = 0,15 mol
nCuO = 3,2 : 80 = 0,04 mol
Theo pt ta thấy CuO pứ hết, H2SO4 dư.
Ta lập tỉ lệ số mol theo pt:
=> mCuO pứ= 3,2 gam
mCuSO4= 0,04 x 160 = 6,4 gam
c/ mdung dịch thu đc = 3,2 + 150 = 153,2 gam
nH2SO4 dư= 0,15 - 0,04 = 0,11 mol
=> mH2SO4 dư = 0,11 x 98 = 10,78 gam
=> C%H2SO4= 10,78 / 153,2 x 100% = 7,03%
C%CuSO4= 6,4 / 153,2 x 100% = 4,18%
*Tk
nCuO=3,2/80=0,04(mol)
PTHH: CuO + H2SO4 -> CuSO4 + H2O
nH2SO4= nCuSO4=nCuO=0,04(mol)
=> mH2SO4=0,04. 98=3,92(g)
=> mddH2SO4= 3,92 : 20%=19,6(g)
mddCuSO4= mCuO + mddH2SO4= 3,2+ 19,6= 22,8(g)
mCuSO4=0,04. 160= 6,4(g)
=> C%ddCuSO4= (6,4/22,8).100=28,07%
nguồn :hoc24
nCuO=3,2/80=0,04(mol)
PTHH: CuO + H2SO4 -> CuSO4 + H2O
nH2SO4= nCuSO4=nCuO=0,04(mol)
=> mH2SO4=0,04. 98=3,92(g)
=> mddH2SO4= 3,92 : 20%=19,6(g)
mddCuSO4= mCuO + mddH2SO4= 3,2+ 19,6= 22,8(g)
mCuSO4=0,04. 160= 6,4(g)
=> C%ddCuSO4= (6,4/22,8).100=28,07%
PTHH: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{Ba\left(OH\right)_2}=0,05\cdot0,5=0,025\left(mol\right)\\n_{HCl}=0,15\cdot0,1=0,015\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,025}{1}>\dfrac{0,015}{2}\) \(\Rightarrow\) Ba(OH)2 còn dư, dd sau p/ứ có tính kiềm
\(\Rightarrow\left\{{}\begin{matrix}n_{BaCl_2}=0,0075\left(mol\right)\\n_{Ba\left(OH\right)_2\left(dư\right)}=0,0175\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{BaCl_2}}=\dfrac{0,0075}{0,05+0,15}=0,0375\left(M\right)\\C_{M_{Ba\left(OH\right)_2}}=\dfrac{0,0175}{0,2}=0,0875\left(M\right)\end{matrix}\right.\)
$n_{Fe_2O_3} = 0,05(mol)$
$n_{H_2SO_4} = \dfrac{150.20\%}{98} = \dfrac{15}{49}(mol)$
$Fe_2O_3 + 3H_2SO_4 \to Fe_2(SO_4)_3 + 3H_2O$
Ta thấy :
$n_{Fe_2O_3} : 1 < n_{H_2SO_4} :3$ nên $H_2SO_4$ dư
$m_{dd\ sau\ pư} = 8 + 150 = 158(gam)$
$n_{H_2SO_4\ dư} = \dfrac{15}{49} - 0,05.3 = \dfrac{153}{980}(mol)$
$n_{Fe_2(SO_4)_3} = 0,025(mol)$
$C\%_{H_2SO_4} = \dfrac{ \dfrac{153}{980}.98}{158} .100\% = 9,7\%$
$C\%_{Fe_2(SO_4)_3} = \dfrac{0,025.400}{158}.100\% = 6,3\%$